Question

Cho a+b+c = 1

Cmr : \(\dfrac{a}{1+b-a}+\dfrac{b}{1+c-b}+\dfrac{c}{1+a-c}\ge1\)

Answer 1

\(\dfrac{a}{1+b-a}+\dfrac{b}{1+c-b}+\dfrac{c}{1+a-c}\)

\(=\dfrac{a}{2b+c}+\dfrac{b}{2c+a}+\dfrac{c}{2a+b}\)

\(=\dfrac{a^2}{2ab+ca}+\dfrac{b^2}{2bc+ab}+\dfrac{c^2}{2ca+bc}\)

\(\ge\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\)

Answer 2

\(\dfrac{a}{1+b-a}+\dfrac{b}{1+c-b}+\dfrac{c}{1+a-c}=\dfrac{a^2}{a+ab-a^2}+\dfrac{b^2}{b+bc-b^2}+\dfrac{c^2}{c+ca-c^2}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c-\left(a^2+b^2+c^2-ab-bc-ca\right)}=\dfrac{1}{1-\left(a^2+b^2+c^2-ab-bc-ca\right)}\ge1\)

Answer 3

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Mà cx chẳng ai giúp đc nx,anh Hung Nguyen thì ít onl